Loops Practice Questions (AP CSA)

(E)	Compile time error

Explanation:

The loop variable i is defined inside the loop and hence is NOT visible outside the loop. Therefore, the code snippet will result in compile-time error.

(A)	0 times

Explanation:

The values of i will run from 1 to 10 and values of j run from 10 down to 1. However, at no point of time they will be equal. Hence, “Hello” will NOT be printed at all.

(B)	Exactly 1 time

Explanation:

The values of i will run from 0 to 9 (i = 10 will violate the loop condition) and values of j run from 10 down to 1. However, the value of i and j will be equal at 5. At this value, the second boolean condition ((i%2 == 1)) will also be satisfied. Hence “Good Morning” will be printed only once.

(D)	Odd
    Odd

Explanation:

We need to run the values of i and j to get the result. However, we can easily make out that EVEN will never be printed because i will never take a even value (values of i start at 1 and are incremented by 2).

(A)	2 times

Explanation:

We need to run the values of k and i to get the result.

(B) 2 3 4 6 6 8 9 10 12 12 14 15

Explanation:

The code prints all multiples of 2 (value of j) and 3 (value of k) between 1 and 15. Moreover, numbers 6 and 12 get printed twice because they are multiples of both 2 and 3.

(C)	Value of i is: 13
    Value of j is: 22

Explanation:

Value of i will be 13 after the loop execution completes.We need to calculate the value of j using the following table:

(E)	Infinite loop 

Explanation:

The code segment results in an infinite loop since the variable value is not been incremented. The condition if (true) always evaluates to true and hence the else part never gets executed and value is not incremented.

(A)	4 8 12 
    8 16 24  

Explanation:

We can easily calculate the output by executing the two nested loops for different values of i and j, as shown below:

(C)	1

Explanation:

The code segment does NOT result in a compile-time error. Although the initialization in the inner for-loop is empty; it is allowed. Additionally, the value of j does not get initialised to 1 every time the inner loop is entered. Hence the output will be 1. Same is shown in table below:

(B) *** [3 times]

Explanation:

We need to trace the values of i and j across the two loops to get the output, as shown below:

(A)	05
    005
    0005

Explanation:

We need to trace the values of k and j across the two loops to get the output, as shown below:

(D)	9600

Explanation:

We need to trace the values of the variables month and salary across the loop to get the output. However, it is important to note here that the code is NOT actually calculating the salary for all the 12 months because the loop variable month is getting incremented twice (once after the execution of the if-else statement and once in the loop execution).

(C)	Prints table of 10 to 100 (all multiples from 1 to 10)

Explanation:

We need to trace the values of the variables i and j across the loops to get the output. The output will be:

10 20 30 40 50 60 70 80 90 100

20 40 60 80 100 120 140 160 180 200

30 60 90 120 150 180 210 240 270 300

40 80 120 160 200 240 280 320 360 400

50 100 150 200 250 300 350 400 450 500

60 120 180 240 300 360 420 480 540 600

70 140 210 280 350 420 490 560 630 700

80 160 240 320 400 480 560 640 720 800

90 180 270 360 450 540 630 720 810 900

100 200 300 400 500 600 700 800 900 1000

(B) for (int i=1, j=2 ; i <= 100; i++)
	{
		if (i%j == 0) System.out.print(i + " ");
	}

Explanation:

Options (C), (D) and (E) will result in printing all odd numbers between 1 and 100. We are now left with option (A) and (B). Option (A) is incorrect because it checks for numbers which when mod with 2 give 2. There are no such numbers. All even numbers when mod with 2 give 0 and all odd numbers give 1. Hence for option (A) nothing gets prints. Option (B) is the correct option, since it checks for numbers which when mod with j (=2) result in a 0.